30 Jun 2025
Angular Momentum, Spin and Symmetry
Angular-momentum algebra, Pauli spinors, central fields, Clebsch-Gordan coupling, conservation laws, parity, identical particles, and spin-orbit coupling.
Angular momentum is the generator of rotations. Its components satisfy
\[[J_i,J_j]=i\hbar\epsilon_{ijk}J_k, \qquad [J^2,J_i]=0.\]Choose simultaneous eigenvectors of $J^2$ and $J_z$:
\[J^2\lvert jm\rangle=\hbar^2j(j+1)\lvert jm\rangle, \qquad J_z\lvert jm\rangle=\hbar m\lvert jm\rangle.\]Eigenvalues from ladder operators
Define $J_\pm=J_x\pm iJ_y$. The commutators give
\[[J_z,J_\pm]=\pm\hbar J_\pm,\]so $J_\pm\lvert jm\rangle$ has magnetic quantum number $m\pm1$. Also
\[J_-J_+=J^2-J_z^2-\hbar J_z, \qquad J_+J_-=J^2-J_z^2+\hbar J_z.\]Taking norms,
\[\left\lVert J_+\lvert jm\rangle\right\rVert^2 =\hbar^2[j(j+1)-m(m+1)]\ge0,\] \[\left\lVert J_-\lvert jm\rangle\right\rVert^2 =\hbar^2[j(j+1)-m(m-1)]\ge0.\]A finite ladder has a top $m_{\max}$ annihilated by $J_+$ and a bottom $m_{\min}$ annihilated by $J_-$. Hence
\[j(j+1)=m_{\max}(m_{\max}+1) =m_{\min}(m_{\min}-1),\]which gives $m_{\max}=j$, $m_{\min}=-j$, and $2j\in{0,1,2,\ldots}$. Therefore
\[\boxed{j=0,\frac12,1,\frac32,\ldots, \qquad m=-j,-j+1,\ldots,j,}\] \[\boxed{ J_\pm\lvert jm\rangle =\hbar\sqrt{j(j+1)-m(m\pm1)}\,\lvert j,m\pm1\rangle. }\]Electron spin and Pauli matrices
For spin $1/2$, $\mathbf S=(\hbar/2)\boldsymbol\sigma$, where
\[\sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad \sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\quad \sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.\]They obey
\[\sigma_i\sigma_j=\delta_{ij}I+i\epsilon_{ijk}\sigma_k, \qquad [S_i,S_j]=i\hbar\epsilon_{ijk}S_k.\]The $S_z$ eigenvectors are
\[\lvert +z\rangle=\begin{pmatrix}1\\0\end{pmatrix}, \qquad \lvert -z\rangle=\begin{pmatrix}0\\1\end{pmatrix},\]with eigenvalues $\pm\hbar/2$. The normalized eigenvectors of $S_x$ are
\[\lvert\pm x\rangle=\frac1{\sqrt2} \begin{pmatrix}1\\\pm1\end{pmatrix},\]and those of $S_y$ are
\[\lvert\pm y\rangle=\frac1{\sqrt2} \begin{pmatrix}1\\\pm i\end{pmatrix}.\]Motion in a centrally symmetric field
For
\[H=\frac{\mathbf p^2}{2\mu}+V(r),\]rotational symmetry gives $[H,\mathbf L]=0$ and $[H,L^2]=0$. For $\psi(\mathbf r)=R_l(r)Y_l^m(\theta,\phi)$, the spherical Laplacian gives
\[\nabla^2(R_lY_l^m) =\left[ \frac1{r^2}\frac d{dr}\left(r^2\frac{dR_l}{dr}\right) -\frac{l(l+1)}{r^2}R_l \right]Y_l^m.\]Set $u_l=rR_l$. Since
\[r^2\frac{dR_l}{dr}=ru_l'-u_l, \qquad \frac1{r^2}\frac d{dr}(ru_l'-u_l)=\frac{u_l''}{r},\]division of the Schrödinger equation by $Y_l^m/r$ gives
\[\boxed{ -\frac{\hbar^2}{2\mu}\frac{d^2u_l}{dr^2} +\left[V(r)+\frac{\hbar^2l(l+1)}{2\mu r^2}\right]u_l=Eu_l. }\]For a nonsingular bound-state solution, $u_l(0)=0$; normalizability requires $u_l(r)\to0$ as $r\to\infty$. The central potential makes the energy independent of $m$.
Addition of angular momenta
For $\mathbf J=\mathbf J_1+\mathbf J_2$, the uncoupled and coupled bases are related by Clebsch-Gordan coefficients:
\[\lvert jm\rangle =\sum_{m_1,m_2} \lvert j_1m_1\rangle\lvert j_2m_2\rangle \langle j_1m_1;j_2m_2\mid jm\rangle.\]Since $J_z=J_{1z}+J_{2z}$, a coefficient can be nonzero only if $m=m_1+m_2$. Ladder termination gives the angular-momentum selection rule
\[\boxed{\lvert j_1-j_2\rvert\le j\le j_1+j_2}\]in integer steps. For two spin-$1/2$ particles,
\[\lvert1,1\rangle=\lvert++\rangle, \qquad \lvert1,0\rangle=\frac{\lvert+-\rangle+\lvert-+\rangle}{\sqrt2},\] \[\lvert1,-1\rangle=\lvert--\rangle, \qquad \lvert0,0\rangle=\frac{\lvert+-\rangle-\lvert-+\rangle}{\sqrt2}.\]The triplet is symmetric under particle exchange; the singlet is antisymmetric.
Space-time symmetries and conservation laws
An infinitesimal continuous transformation generated by $G$ is $U(\varepsilon)=I-i\varepsilon G/\hbar$. If the Hamiltonian is invariant, $UHU^\dagger=H$, then to first order
\[H-\frac{i\varepsilon}{\hbar}[G,H]=H \quad\Longrightarrow\quad [G,H]=0.\]The Heisenberg equation then gives $d\langle G\rangle/dt=0$. Spatial translations are generated by $\mathbf P$, rotations by $\mathbf J$, and time translations by $H$ itself. Their invariance therefore conserves linear momentum, angular momentum, and energy, respectively.
Parity is the discrete inversion
\[\Pi\psi(\mathbf r)=\psi(-\mathbf r), \qquad \Pi^2=I.\]Its eigenvalues are $\pm1$. For a parity-even Hamiltonian, $[H,\Pi]=0$, so parity is conserved. Central-field states have parity $(-1)^l$ because $Y_l^m(-\hat{\mathbf r})=(-1)^lY_l^m(\hat{\mathbf r})$.
Exclusion, exchange symmetry, and spin-orbit coupling
For identical particles, exchanging labels cannot change any observable. For distinct orthonormal one-particle orbitals $\phi_a$ and $\phi_b$, the normalized symmetric and antisymmetric combinations are
\[\Psi_\pm(1,2)=\frac1{\sqrt2} [\phi_a(1)\phi_b(2)\pm\phi_b(1)\phi_a(2)].\]Electrons are fermions, so the complete state must be antisymmetric. If the two orbital labels coincide, applying the antisymmetrizer gives zero rather than a normalizable state; the symmetric product is simply $\phi_a(1)\phi_a(2)$. Thus no two electrons may occupy the same complete one-particle state. A symmetric spin triplet must accompany an antisymmetric spatial state, while the antisymmetric spin singlet accompanies a symmetric spatial state.
With electron spin $s=1/2$, total angular momentum is $\mathbf J=\mathbf L+\mathbf S$. The identity
\[J^2=L^2+S^2+2\mathbf L\cdot\mathbf S\]gives
\[\boxed{ \mathbf L\cdot\mathbf S\,\lvert lsjm\rangle =\frac{\hbar^2}{2} [j(j+1)-l(l+1)-s(s+1)]\lvert lsjm\rangle. }\]For $l>0$, the allowed values are $j=l\pm1/2$. Therefore
\[\langle\mathbf L\cdot\mathbf S\rangle =\begin{cases} (\hbar^2/2)l,&j=l+1/2,\\ -(\hbar^2/2)(l+1),&j=l-1/2. \end{cases}\]For an electron moving in a central potential energy $V(r)$, the nonrelativistic spin-orbit interaction, including the Thomas factor, is
\[\boxed{ H_{SO}=\frac1{2m_e^2c^2r}\frac{dV}{dr}\, \mathbf L\cdot\mathbf S. }\]It consequently splits a fixed $l$ level into the two allowed total-angular-momentum levels $j=l\pm1/2$.
For a fixed radial state, define the radial expectation
\[\boxed{ \xi_{nl}=\left\langle nl\right\lvert \frac1{2m_e^2c^2r}\frac{dV}{dr} \left\lvert nl\right\rangle. }\]The first-order spin-orbit shift is therefore
\[\boxed{ \Delta E_{nlj} =\frac{\xi_{nl}\hbar^2}{2} [j(j+1)-l(l+1)-s(s+1)]. }\]The level ordering in the figure assumes $\xi_{n1}>0$; a negative radial expectation reverses the two levels.
The Pauli algebra, central-field radial substitution, singlet-triplet normalization, and spin-orbit eigenvalues are verified in the Maxima worksheet; every printed residual is zero.
Discussion